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Architecture of GP sums

Authors
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    Name
    Vu Hung
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Problem Statement

Let a GP have first term a≠0a\neq0, ratio r≠0,±1r\neq 0,\pm1, and partial sums SnS_n.

  • Prove S2nSn=rn+1\dfrac{S_{2n}}{S_n}=r^n+1.
  • Given S12=65S6S_{12}=65S_6 and second term ar=12ar=12, find possible (a,r)(a,r).
  • If Σn\Sigma_n is the partial-sum sequence of a GP with first term aa and ratio r2r^2, show
ΣnSn=rn+1r+1.\frac{\Sigma_n}{S_n}=\frac{r^n+1}{r+1}.

Hints

Use Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1} and difference-of-squares factorization.

Solutions

S2nSn=r2n−1rn−1=rn+1.\frac{S_{2n}}{S_n}=\frac{r^{2n}-1}{r^n-1}=r^n+1.

So r6+1=65  ⟹  r6=64  ⟹  r=±2r^6+1=65\implies r^6=64\implies r=\pm2. Since ar=12ar=12, possible pairs are (a,r)=(6,2)(a,r)=(6,2) or (−6,−2)(-6,-2). Also

Σn=a(r2n−1)r2−1,\Sigma_n=\frac{a(r^{2n}-1)}{r^2-1},

thus

ΣnSn=r2n−1(r+1)(rn−1)=rn+1r+1.\frac{\Sigma_n}{S_n} =\frac{r^{2n}-1}{(r+1)(r^n-1)} =\frac{r^n+1}{r+1}.

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